Chapter 2, Problem 81EP

(a).

To determine

To find: The velocity of the particle at any time $t$ .

Answer to Problem 81EP

The velocity is $v\left(t\right)=3t^2-12$ .

Explanation of Solution

Given:

  $x\left(t\right)=t^3-12t+5$ at $t\ge 0$

Concept Used:

Velocity is the derivative of the positive function with respect to time. At time $t$ the velocity is:

  $v\left(t\right)=\frac{ds}{dt}=\underset{\Delta t\to 0}{\lim }\frac{f\left(t+\Delta t\right)-f\left(t\right)}{\Delta t}$

Calculation:

  $\begin{array}{c}v\left(t\right)=\frac{d}{dt}\left[x\left(t\right)\right]\\ =\frac{d}{dt}(t^3-12t+5)\\ =\frac{d}{dt}\left[t^3\right]-12\cdot \frac{d}{dt}\left[t\right]+\frac{d}{dt}\left[5\right]\\ =3t^2-12\end{array}$

Conclusion:

The velocity is $v\left(t\right)=3t^2-12$ .

b.

To determine

To calculate: The acceleration at $t$

Answer to Problem 81EP

The acceleration is $a\left(t\right)=6t$ .

Explanation of Solution

Given:

  $x\left(t\right)=t^3-12t+5$

Concept Used:

The acceleration is the derivative of velocity with respect to time. If velocity at time $t$ is $v(t)=\frac{ds}{dt}$ , then the acceleration at time $t$ is:

  $a\left(t\right)=\frac{dv}{dt}=\frac{d^{2}s}{d{t}^2}$

Calculation:

  $\begin{array}{c}a\left(t\right)=\frac{dv}{dt}\\ =\frac{d}{dt}\left[3t^2-12\right]\\ =3\cdot \frac{d}{dt}\left[t^2\right]-\frac{d}{dt}\left[12\right]\\ =3\left[2t\right]-0\\ =6t\end{array}$

Conclusion:

The acceleration is $a\left(t\right)=6t$ .

c.

To determine

To calculate: All values of $t$ for which the particle is at rest .

Answer to Problem 81EP

The value of $t=2$ .

Explanation of Solution

Given:

  $x\left(t\right)=t^3-12t+5$

Concept Used:

For the value of $t$ for which the particle is at rest is:

  $\frac{dx(t)}{dt}=0$

Calculation:

  $\begin{array}{l}\frac{d}{dt}\left[x\left(t\right)\right]=0\\ 3t^2-12=0\\ 3t^2=12\\ t^2=4\\ t=2\end{array}$

Conclusion:

The value of $t=2$ .

d.

To determine

Find the speed of the particle when its acceleration is zero

Answer to Problem 81EP

The speed of the particle is $-12$ when its acceleration is zero.

Explanation of Solution

Given:

  $x\left(t\right)=t^3-12t+5$

Concept Used:

The acceleration is the derivative of velocity with respect to time. If velocity at time $t$ is $v(t)=\frac{ds}{dt}$ , then the acceleration at time $t$ is:

  $a\left(t\right)=\frac{dv}{dt}=\frac{d^{2}s}{d{t}^2}$

Calculation:

  $\begin{array}{l}\frac{d^{2}x}{d{t}^2}=0\\ 6t=0\\ t=0\end{array}$

Velocity at $t=0$

  $\begin{array}{c}{\left[\frac{d}{dt}\left[x\left(t\right)\right]\right]}_{t=0}=3{\left(t\right)}^2-12\\ =-12\end{array}$

Conclusion:

The speed of the particle is $-12$ when its acceleration is zero.

e.

To determine

Find that if the particle moving toward the origin or away from the origin when $t=3$ .

Answer to Problem 81EP

The particle is moving towards origin.

Explanation of Solution

Given:

  $x\left(t\right)=t^3-12t+5$

Concept Used:

Velocity is the derivative of the positive function with respect to time. At time $t$ the velocity is:

  $v\left(t\right)=\frac{ds}{dt}=\underset{\Delta t\to 0}{\lim }\frac{f\left(t+\Delta t\right)-f\left(t\right)}{\Delta t}$

Calculation:

At $t=3$ :

  $\begin{array}{c}x\left(3\right)=3^3-12\times 3+5\\ =27-36+5\\ =-4\\ {\left[\frac{dx(t)}{dt}\right]}_{t=3}=3\cdot {\left(3\right)}^2-12\\ =27-12\\ =15\end{array}$

Conclusion:

At $t=3$ the position of the particle is negative, and velocity is positive.

Therefore, the particle is moving towards origin.